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ProblemEditorialSubmissionsComments

Maximum Occuring Character

BasicAccuracy: 50.0%Submissions: 34093Points: 1

Given a string str. The task is to find the maximum occurring character in the string str. If more than one character occurs the maximum number of time then print the lexicographically smaller character.

Example 1:

Input:
str = testsample
Output: e
Explanation: e is the character which
is having the highest frequency.

Example 2:

Input:
str = output
Output: t
Explanation:  t and u are the characters
with the same frequency, but t is
lexicographically smaller.

Your Task:
The task is to complete the function getMaxOccuringChar() which returns the character which is most occurring.

Expected Time Complexity: O(N).
Expected Auxiliary Space: O(Number of distinct characters).
Note: N = |s|

Constraints:
1 ≤ |s| ≤ 100

View Bookmarked Problems
Company Tags
Flipkart
Topic Tags
Strings
Related Courses
Data Structures and Algorithms - Self PacedComplete Interview Preparation - Self PacedFlipkart SDE Preparation Test SeriesDSA Live for Working Professionals - LiveCompetitive Programming - LiveComplete Test Series for Product-Based Companies
Related Interview Experiences
Flipkart Interview Set 2 Sde 2
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Output Window

Compilation ResultsCustom Input

Problem Solved Successfully

Test Cases Passed:
50 /50
Total Points Scored:
1 /1
Your Total Score:
1
Total Time Taken:
0.01
Your Accuracy:
100%
Attempts No.:
1

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class Solution
{
public:
//Function to find the maximum occurring character in a string.
char getMaxOccuringChar(string str)
{
int arr[26] = {0};
for(int i=0; i<str.length(); i++){
char ch = str[i];
int number = 0;
if(ch >= 'a' && ch <= 'z'){
number = ch - 'a';
}
else{
number = ch - 'A';
}
arr[number]++;
}
int maxi = -1, ans= 0;
for(int i=0; i<26; i++){
if(maxi < arr[i]){
ans = i;
maxi = arr[i];
}
}
char finalAns = 'a' + ans;
 
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Output Window

Compilation ResultsCustom Input

Problem Solved Successfully

Test Cases Passed:
50 /50
Total Points Scored:
1 /1
Your Total Score:
1
Total Time Taken:
0.01
Your Accuracy:
100%
Attempts No.:
1
If you are facing any issue on this page. Please let us know.